Undamped Free Vibration MCQ Quiz - Objective Question with Answer for Undamped Free Vibration - Download Free PDF

Explanation:

Flexible Shaft

• A flexible shaft is a type of mechanical component that is specifically designed to transmit rotary motion and torque efficiently between two points that may not be perfectly aligned. It consists of a flexible core that allows it to bend and twist, making it suitable for applications where rigid shafts would be impractical or impossible to use.
• The flexible shaft is made up of a series of tightly wound helical coils, which provide the necessary flexibility and strength to transmit torque. These coils allow the shaft to bend and adapt to changes in alignment while maintaining its ability to rotate. The design ensures low bending rigidity while maintaining relatively high torsional rigidity, allowing the shaft to twist without significant loss of torque transmission.

Advantages:

• Ability to transmit motion and torque around obstacles or in confined spaces.
• Lightweight and compact design compared to rigid shafts.
• High torsional strength while being flexible in bending.
• Reduces the need for precise alignment of components.

Disadvantages:

• Not suitable for applications requiring extremely high torque or bending loads.
• May experience wear and tear over time due to constant bending and twisting.
• Limited in length and load-carrying capacity compared to rigid shafts.

Applications: Flexible shafts are commonly used in applications such as power tools, automotive speedometers, dental equipment, and various industrial machines where motion needs to be transmitted in a flexible and adaptable manner.

Concept:

When a mechanical system is subjected to harmonic excitation, the response of the system is influenced by the excitation frequency and the damping present.

The ratio of the amplitude of the system's response to the static deflection under the same force is called the Dynamic Magnification Factor (DMF).

General Formula:

\( \text{DMF} = \frac{1}{\sqrt{(1 - r^2)^2 + (2\xi r)^2}} \)

Where,
\( r = \frac{\omega}{\omega_n} \) is the frequency ratio
\( \xi \) is the damping ratio.

At Resonance:

At resonance, excitation frequency equals natural frequency, i.e., \( \omega = \omega_n \Rightarrow r = 1 \).

Substitute in the equation:

\( \text{DMF}_{\text{resonance}} = \frac{1}{\sqrt{(1 - 1)^2 + (2\xi)^2}} = \frac{1}{2\xi} \)

Concept:

For a simply supported shaft with a rotor (concentrated mass) at the center, the critical (whirling) speed is determined using the relation:

\( \omega = \sqrt{\frac{g}{\delta}} \)

Where, \( \delta \) is the static deflection due to the weight of the rotor.

Static deflection at center of simply supported shaft due to point load is given by:

\( \delta = \frac{W L^3}{48 EI} \)

Calculation:

Given:

Mass of rotor, \(m = 12~kg, so ~weight, W = mg = 12 \times 10 = 120~N\)

Span of shaft, \(L = 1~m, ~EI = 4~kN\cdot m^2 = 4000~N\cdot m^2\)

Substitute in deflection formula:

\( \delta = \frac{120 \times 1^3}{48 \times 4000} = \frac{120}{192000} = \frac{1}{1600}~m \)

Now, critical speed:

\( \omega = \sqrt{\frac{g}{\delta}} = \sqrt{10 \times 1600} = \sqrt{16000} = 40\sqrt{10}~rad/s \)

Concept:

The energy absorbed by a spring is given by the formula:

\(U = \frac{1}{2} k x^2\)

Where, U is the energy stored, k is the stiffness of the spring, and x is the deflection from its natural (free) length.

Calculation:

To increase the energy absorption capacity without changing the stiffness k, we need to increase the value of x, i.e., the maximum allowable deflection of the spring.

By increasing the spring's free length, the spring can be deflected more before reaching its solid height (fully compressed condition), allowing more energy to be absorbed.

Concept:

Springs in series:

\(\frac{1}{{{k_e}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} \)

Springs in parallel:

ke = k1 + k2

Calculation:

Given:

Springs with spring constant k and k are connected in parallel,

Let their equivalent spring constant be k1

k1 = k + k

k1 = 2k

Now k1, k and k are in series connection, let their equivalent spring constant is keq

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{k_e}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{2k}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{k_{eq}}=\frac{1+2+2}{2k}\)

\(\frac{1}{k_{eq}}=\frac{5}{2k}\)

\(k_{eq}=\frac{2k}{5}\)

k_eq = 0.4 k

Concept:

Springs in series:

\(\frac{1}{{{k_e}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} \)

Springs in parallel:

ke = k1 + k2

Calculation:

Given:

Springs with spring constant k and k are connected in parallel,

Let their equivalent spring constant be k1

k1 = k + k

k1 = 2k

Now k1, k and k are in series connection, let their equivalent spring constant is keq

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{k_e}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{2k}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{k_{eq}}=\frac{1+2+2}{2k}\)

\(\frac{1}{k_{eq}}=\frac{5}{2k}\)

\(k_{eq}=\frac{2k}{5}\)

k_eq = 0.4 k

Concept:

\(Time~ Period= T =2\pi √{\frac Lg}\)

where L = Length of the pendulum, g = acceleration due to gravity

Calculation:

Given:

at moon 

\(g'=\frac g6\)

where g' = acceleration due to gravity at the moon

\(New~Time~ Period= T' =2\pi √ {\frac L{g'}}=2\pi √ {\frac {6L}g}\)

 \(T'=√ {6}T\)

As time period for oscillation increases √6 times, therefore speed becomes √6 times slower.

Concept:

When spring is connected in parallel as shown, the equivalent stiffness is the sum of all individual stiffness of spring.

k_eq = k₁ + k₂

The natural frequency ω_n of a spring-mass system is given by:

\(\omega_n=\sqrt{\frac{k_{eq}}{m}}\;\;\; and\;\;\;\omega_n=2\pi f\)

k_eq = equivalent stiffness and m = mass of body.

Calculation:

Given:

k₁ = 4000 N/m, k₂ = 1600 N/m and m = 1.4 kg

Springs are in parallel correction

∴ k_eq = k₁ + k₂ = 4000 + 1600 = 5600 N/m

\(ω_n = \sqrt {\frac{k_{eq}}{{m}}}\Rightarrow \sqrt {\frac{{5600}}{{1.4}}} = \sqrt {4000} \)

\(f = \frac{ω_n}{{2\pi }} \Rightarrow \frac{{\sqrt {4000} }}{{2\pi }} \approx 10\;Hz\)

Explanation:-

Critical or whirling speed of a shaft

• When the rotational speed of the system coincides with the natural frequency of lateral/transverse vibrations, the shaft tends to bow out with a large amplitude. This speed is termed as critical/whirling speed.
• Whirling speed or Critical speed of a shaft is defined as the speed at which a rotating shaft will tend to vibrate violently in the transverse direction if the shaft rotates in the horizontal direction.
• In other words, the whirling or critical speed is the speed at which resonance occurs.
• Hence we can say that Whirling of the shaft occurs when the natural frequency of transverse vibration matches the frequency of a rotating shaft.
• It is the speed at which the shaft runs so that the additional deflection of the shaft from the axis of rotation becomes infinite is known as critical or whirling speed.

Deflection of the shaft due to transverse vibration of the shaft.

\(y = \frac{e}{{{{\left( {\frac{{{\omega _n}}}{\omega }} \right)}^2} - 1\;}}\)

At critical speed,

\(\omega = {\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{g}{\delta }} \)

where,

ω = Angular velocity of the shaft, k = Stiffness of shaft, e = initial eccentricity of the center of mass of the rotor

m = mass of rotor, y = additional of rotor due to centrifugal force.

Concept:

Transmissibility is given by:

\(\varepsilon = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left[ {2\zeta \frac{\omega }{{{\omega _n}}}} \right]}^2}} }}\)

\(C = 2\zeta \sqrt {KM} \Rightarrow \zeta = \frac{C}{{2\sqrt {KM} }}\)

Calculation:

After additional damper in parallel \(\zeta ' = \frac{{C + C'}}{{2\sqrt {KM} }}\)

Thus  \(\zeta ' > \zeta \)

Now,  \({\omega _d} = {\omega _n}\sqrt {1 - {\zeta ^2}} \)  

 \(\zeta ' > \zeta \)indicating that \(\omega _d' < {\omega _d}\)

\(time\ period\left( {{T_d}} \right) = \frac{{2\pi }}{{{\omega _d}}}\)

Concept:

When both rotors are free to oscillate the node will be formed at the center (when the rotor is similar) but when one of the rotors is fixed the node will shift to the fixed end.

Calculation:

In the two-rotor system shown in figure

ω₁ = ω₂

\(\sqrt {\frac{{{K_{t1}}}}{{{J_1}}}} = \sqrt {\frac{{{K_{t2}}}}{{{J_2}}}} \)

\({K_t} = \frac{{G{I_p}}}{L}\)

where, \({K_t}\) = torsional stiffness

We have, depending on the position of the node effective length will be

• equal to L/2 when both rotors are free to oscillate.
• equal to L when one of the rotors is fixed.

Hence, by fixing the rotor,

• effective length increases.
• K_t value will decrease.
• natural frequency ω_n will decrease.

Explanation:-

Critical or whirling speed of a shaft

• When the rotational speed of the system coincides with the natural frequency of lateral/transverse vibrations, the shaft tends to bow out with a large amplitude. This speed is termed as critical/whirling speed.
• Whirling speed or Critical speed of a shaft is defined as the speed at which a rotating shaft will tend to vibrate violently in the transverse direction if the shaft rotates in the horizontal direction.
• In other words, the whirling or critical speed is the speed at which resonance occurs.
• Hence we can say that whirling of the shaft occurs when the natural frequency of transverse vibration matches the frequency of a rotating shaft.
• It is the speed at which the shaft runs so that the additional deflection of the shaft from the axis of rotation becomes infinite is known as critical or whirling speed.

Explanation:

Vibration

Vibration is a periodic motion of small magnitude. But for sake of simplicity, we can assume it as a simple harmonic motion of small amplitude.

Transverse vibrations

• When the particles of the shaft or disc move approximately perpendicular to the axis of the shaft, then the vibrations are known as transverse vibrations.
• In this case, the shaft is straight and bent alternately and bending stresses are induced in the shaft.
• Rotating shafts tend to vibrate violently in Transverse direction at critical speeds.

Longitudinal vibrations

• When the particles of the shaft or disc move parallel to the axis of the shaft, then the vibrations are known as longitudinal vibrations.
• In this case, the shaft is elongated and shortened alternately and thus the tensile and compressive stresses are induced alternately in the shaft.

Torsional vibrations

• ·When the particles of the shaft or disc move in a circle about the axis of the shaft, then the vibrations are known as torsional vibrations.
• ·In this case, the shaft is elongated and shortened alternately and thus the tensile and compressive stresses are induced alternately in the shaft.

Critical or whirling speed of a shaft

The speed, at which the shaft runs so that the additional deflection of the shaft from the axis of rotation becomes infinite, is known as critical or whirling speed.

Concept:

The natural frequency is given by:

\(\omega = \sqrt {\frac{k_{eq}}{m}} =\sqrt {\frac{g}{\delta }} \;\)

Calculation:

Given:

L₁ = L, \(L_2=\frac{L_1}{2}\)

For cantilever beam:

\(\begin{array}{l} \delta = \frac{{P{L^3}}}{{3EI}}\\ \omega = \sqrt {\frac{g}{\delta }} = \sqrt {\frac{{3EIg}}{{P{L^3}}}} \Rightarrow \omega \propto \sqrt {\frac{1}{{{L^3}}}} \\ \frac{{{\omega _2}}}{{{\omega _1}}} = \sqrt {\frac{{L_1^3}}{{L_2^3}}} = \sqrt {\frac{{{L^3}}}{{{{\left( {\frac{L}{2}} \right)}^3}}}} = 2\sqrt2 \\ \therefore\;{\omega _2} > {\omega _1} \end{array}\)

Concept

Critical speed

Whirling speed or Critical speed of a shaft is defined as the speed at which a rotating shaft will tend to vibrate violently in the transverse direction if the shaft rotates in horizontal direction. In other words, the whirling or critical speed is the speed at which resonance occurs.

Critical speed of the shaft:

\(\omega = \sqrt {\frac{g}{{\rm{Δ}}}} \)

Calculation:

Given:

Δ =  25 mm

\(\omega = \sqrt {\frac{g}{{\rm{Δ}}}} = \sqrt {\frac{{10}}{{25 \times {{10}^{ - 3}}}}} = \sqrt {\frac{{10000}}{25}} =\sqrt {400} =20 \)  rad/sec